Central force and acceleration in the polar direction

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SUMMARY

The discussion centers on the behavior of a particle under a central force, defined as a radial force represented by ##\vec{F}=f(r) \hat{r}##. It establishes that while the radial force does not exert torque, angular momentum remains conserved, leading to a constant product of ##r^2 v_{\theta}##. As the radius ##r## decreases, the tangential velocity ##v_{\theta}## must increase, indicating that even in the absence of a tangential force, the magnitude of ##v_{\theta}## can change due to the conservation of angular momentum. The conversation clarifies that the acceleration in the tangential direction remains zero despite changes in velocity magnitude.

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  • Understanding of central forces and their properties
  • Familiarity with angular momentum conservation principles
  • Knowledge of polar coordinates and their application in physics
  • Basic concepts of motion in non-inertial reference frames
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  • Explore the equations of motion for particles in polar coordinates
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Soren4
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Consider a central force. The central force is radial by definition, so ##\vec{F}=f(r) \hat{r}##. Therefore, by definition, the acceleration caused by the force, in the direction of ##\hat{\theta}## must be zero, ##\vec{a_{\theta}}=0##.

In presence of central force angular momentum is conserved, in particular its magnitude, which is ##|\vec{L}|=m r^2 v_{\theta}##, where ##v_{\theta}## is the magnitude of the velocity of the body in the direction of ##\hat{\theta}##.
That means that the product ##r^2 v_{\theta}## stays constant and, if ##r## gets smaller, ##v_{\theta}## must increase.

Since ##v_{\theta}## is the magnitude of the velocity of the body in the direction of ##\hat{\theta}##, this means that there must be and acceleration in this direction, necessarily.

But how can this be possible, since, as said, the force is completely radial?

I'm ok with the fact that the force is not orthogonal to the velocity, hence the magnitude of the velocity can increase, but, in particular, I really don't see how the magnitude of the component ##v_{\theta}## can change, provided the fact that the force is radial.

In this picture it is clear that the magnitude of the velocity (which is tangential to the ellipse, of course) changes, but I can't understand the change in ##v_{\theta}##.

https://www.physicsforums.com/attachments/99029
 
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No, it doesn't mean that there is a theta component. It means that the initial velocity is v_theta. If there was an acceleration along theta, angular momentum wouldn't be conserved. There would be a non-zero net torque.
 
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What does v equal in polar coordinates?
 
Soren4 said:
I'm ok with the fact that the force is not orthogonal to the velocity, hence the magnitude of the velocity can increase, but, in particular, I really don't see how the magnitude of the component ##v_{\theta}## can change, provided the fact that the force is radial.
Before you take on elliptical orbits in a central force, consider the simpler case of no force at all. What is the equation of motion of a particle that starts at the point ##(\theta=0,r=1))## with a speed ##v## in the tangential direction? It's not subject to any force or acceleration at all, yet neither the ##\theta## nor the ##r## components of its velocity are constant.

Cartesian coordinates have the nice property that the basis vectors are functions of neither position nor time, so when you rewrite ##\vec{F}=m\vec{a}## as the differential equation ##\frac{d^2\vec{r}(t)}{dt^2}=\frac{\vec{F}}{m}## it simplifies into ##\frac{d^2A_x(t)}{dt^2}=\frac{F_x}{m}## (and likewise for the y and z components) and you can safely conclude that if ##F_x=0## then ##v_x## will be a constant. More general coordinate systems don't necessarily work that way.
 
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vθ not being constant does not mean that the acceleration in the theta direction is not zero.
 
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